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3 years ago

11

Which of the following is not tru about whorl patterns

1 answer:

tamaranim1 [39]3 years ago

6 0

You didn’t put any problem

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OLga [1]

The age of earth I think

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3 years ago

Read 2 more answers

Why does nuclear charge increase as you go left to right across a period

MariettaO [177]

The atomic number increases moving left to right across a period and subsequently so does the effective nuclear charge. Therefore, moving left to right across a period the nucleus has a greater pull on the outer electrons and the atomic radii decreases.

7 0

3 years ago

Unit 1- Unit Assessment (Matter & Density)

konstantin123 [22]

Answer:

<h3>The answer is 11 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 3025 g

volume = 275 mL

We have

$density = \frac{3025}{275} \\$

We have the final answer as

<h3>11 g/mL</h3>

Hope this helps you

7 0

3 years ago

7. The grid in a triode is kept negatively charged to prevent A. electrons from escaping from the tube containing the triode. B.

romanna [79]

Answer:

A. Electrons from escaping from the tube containing the triode.

:)

Explanation:

5 0

3 years ago

For the simple decomposition reactionAB(g)→ A(g) + B(g)Rate =k[AB]2 and k=0.2 L/mol*s . How long will it takefor [AB] to reach 1

mixer [17]

Answer:

6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.

Explanation:

$Rate = k[AB]^2$

The order of the reaction is 2.

Integrated rate law for second order kinetic is:

$\frac{1}{[A_t]} = \frac{1}{[A]_0}+kt$

Where, $[A_0]$ is the initial concentration = 1.50 mol/L

$[A_t]$ is the final concentration = 1/3 of initial concentration = $\frac{1}{3}\times 1.50\ mol/L$ = 0.5 mol/L

Rate constant, k = 0.2 L/mol*s

Applying in the above equation as:-

$\frac{1}{0.5} = \frac{1}{1.50}+0.2t$

$\frac{1}{1.5}+0.2x=\frac{1}{0.5}$

$t = 6.66\ s$

<u>6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.</u>

5 0

3 years ago